$\lim_{x\to 2}\dfrac{3-\sqrt{5x-1}}{x-2}=$
Substituting $x=2$ into $\dfrac{3-\sqrt{5x-1}}{x-2}$ results in the indeterminate form $\dfrac{0}{0}$. This doesn't necessarily mean the limit doesn't exist, but it does mean we have to work a little before we find it. Since we have a rational expression with a square root on our hands, let's try to re-write it using the method of rationalization. $\begin{aligned} &\phantom{=}\dfrac{3-\sqrt{5x-1}}{x-2} \\\\ &=\dfrac{3-\sqrt{5x-1}}{x-2}\cdot\dfrac{3+\sqrt{5x-1}}{3+\sqrt{5x-1}} \gray{\text{Rationalize the numerator}} \\\\ &=\dfrac{3^2-(5x-1)}{(x-2)(3+\sqrt{5x-1})} \\\\ &=\dfrac{-5\cancel{(x-2)}}{\cancel{(x-2)}(3+\sqrt{5x-1})} \gray{\text{Cancel out common factors}} \\\\ &=\dfrac{-5}{3+\sqrt{5x-1}} \text{, for }x\neq 2 \end{aligned}$ This means that the two expressions have the same value for all $x$ -values (in their domains) except for $2$. We can now use the following theorem: If $f(x)=g(x)$ for all $x$ -values in a given interval except for $x=c$, then $\lim_{x\to c}f(x)=\lim_{x\to c}g(x)$. In our case, $\dfrac{3-\sqrt{5x-1}}{x-2}=\dfrac{-5}{3+\sqrt{5x-1}}$ for all $x$ -values in the interval $(1.5,2.5)$ except for $x=2$. Therefore, $\lim_{x\to 2}\dfrac{3-\sqrt{5x-1}}{x-2}=\lim_{x\to 2}\dfrac{-5}{3+\sqrt{5x-1}}=-\dfrac{5}{6}$. (The last limit was found using direct substitution.) [I want to see how this looks graphically!] In conclusion, $\lim_{x\to 2}\dfrac{3-\sqrt{5x-1}}{x-2}=-\dfrac{5}{6}$.